What is the integral of (12)/(1-cos(4x)) ?

The integral of (12)/(1-cos(4x)) is -3cot(2x)+C

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integral of (12)/(1-cos(4x))

\int \frac{12}{1 - cos ( 4 x )} d x

- 3 cot (2 x) + C

Solution steps

\int \frac{12}{1 - cos ( 4 x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( u ) )} d u

= 12 \cdot \int \frac{1}{4 ( 1 - cos ( u ) )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 12 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 12 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 12 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

12 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - 3 cot (2 x)

= - 3 cot (2 x)

Add a constant to the solution

= - 3 cot (2 x) + C

n = 0 \sum \infty \frac{1}{6}

\int - \frac{3}{2}

f (x) = 2 x^{2}

y (1 + x^{2}) (\frac{d y}{d x}) = x (7 + y^{2})

\frac{d}{d x} (\frac{x}{x + 8})

What is the integral of (12)/(1-cos(4x)) ?
The integral of (12)/(1-cos(4x)) is -3cot(2x)+C